 So, we have a couple of properties of these dual norms. The first is that let f be a pre-norm on r to the n, then mod of y transpose x is less than or equal to f of x f d of y and it is also less than or equal to f d of x times f of y. So, this is a little reminiscent of the Gauss-Schwarz inequality, which says that mod of y transpose x is less than or equal to the L2 norm of x times the L2 norm of y. It turns out that the dual norm of the L2 norm is the L2 norm itself. So, when I consider f to be the L2 norm, which is of course a vector norm, but it is also a pre-norm, then these inequalities reduce to the Gauss-Schwarz inequality and this is true for every x, y belonging to r to the n. But this is a more general inequality. Now, how do you show this? So, of course, if for example, x was equal to 0, then the left hand side is 0. The right hand side is also equal to 0 in either case and so the inequalities certainly holds when x equals 0. So, if x equals 0, so let x be nonzero. Then what we have is let me consider the quantity y transpose x divided by f of x. Since x is not equal to 0, f of x is nonzero. So, I can consider the vector x over f of x and this is less than or equal to the max over all z such that f of z equals 1 of mod of y transpose z. Why is this true? Of course, this is equal to f d of y by definition and inequality immediately follows. So, this implies I am just going to take this f of x to the other side. It is strictly positive. So, mod y transpose x is less than or equal to f of x f d of y. But why is this inequality true? This is trivial. I just made the same argument x over f of x is one such vector such that f of z equals 1. If I took f of x over f of x, I will get the value equal to 1. So, it satisfies this constraint and so here what I am doing on the right hand side is I am not restricting myself to x over f of x. Instead, I am considering all possible z such that f of z equals 1 and I am maximizing y transpose z. So, this is like one candidate solution to this optimization problem and since this is maximizing this quantity, this has to be at least equal to its value at one of the feasible points and so then that is equal to f d of y. Now, the rest of the proof follows immediately because mod of y transpose x is equal to mod of x transpose y and so I can just exchange x and y and then I will have y transpose x magnitude is less than or equal to f d of x times f of y. So, now that we have defined dual norms, we can ask are there some, we looked at some examples of norms and we can ask what are the duals of those norms. So, here is one result which will help to answer that if x and y are vectors in r to the n then mod of y transpose x is equal to mod of sigma i equal to 1 to n y i x i which is less than or equal to I will take the mod inside that will only increase the value or it cannot decrease the value i equal to 1 to n mod of y i x i which is less than or equal to. Now, what I can do is in this summation I can pull out the largest value of y and in magnitude and that will only in other words all these y i's I will replace with the biggest of these biggest of y 1 through y n and so I and then that is just some single number that is multiplying all these x's. So, I will write it like this. So, max 1 less than or equal to i less than or equal to n mod y i times sigma i equal to and let me just write it with j so that you do not get confused j equal to 1 to n mod x j and this and this is equal to by definition the max entry of max modulus entry of y is what we call the infinity norm and the sum of magnitudes of x is what we call the l 1 norm. So, what we have is that y transpose x is less than or equal to y infinity times x l 1. This is like similar to the Cauchy-Schwarz inequality there the two norms being operated with the l 2 norm and the l 2 norm here I have the infinity norm and I have the l 1 norm. This is in fact a special case of of an inequality known as Holder's inequality which says that mod of y transpose x is less than or equal to norm x l p norm times norm y l q norm where p and q are such that 1 over p plus 1 over q equals 1. So, for example, if I choose p equals 1 then I must choose q equals infinity so that 1 plus 1 over infinity which is 0 equals 1. So, it reduces to this inequality when I said p equals 1. So, the question now I can ask is if I am given y when will equality hold here or for what x will equality hold? Okay, then for that you have to examine where we did this inequalities here and ask when will this these inequalities hold with equality. Now, the when you take the modulus inside this will be this will hold with equality if each of these terms were already non-negative and so then there is no cancellations that are happening across the terms here and so taking the mod inside doesn't really change this value so the two will be equal. So, this should be non-negative and then when will this not affect it? It won't affect it if the x is such that pulling out the maximum value of y i doesn't affect this overall summation. In other words, if x was chosen such that it has a non-zero entry only for the entry of y which solves this optimization problem that is suppose the first entry of y was the maximum magnitude entry then if only the first entry of x was non-zero and all other entries of x were equal to zero then the other terms in the summation are not contributing to the sum anyway and so then pulling out this maximum value which is the first entry of y is not going to change the value of value of this quantity I mean these two will become equal so so so basically so so so suppose x was such that. Sir. Yeah. Alternatively we can say that all the values of y are equal will that make sense? No I am saying y is given I will come back to that that's a good point I will come back to this point in a moment when I ask the alternative question which is when given an x when will the equality hold then your answer is absolutely right you want to choose all the values of y to be equal but now I'm looking over all x is such that the L1 norm of x is equal to 1 and I'm asking when will equality hold in in here? Okay sir yeah. In this inequality and so if so the equality holds when xi equals 1 for i equal to the argument 1 less than or equal to k less than or equal to n that maximizes mod yk and zero otherwise then so so that is the x in other words what I've just done is I've actually solved the problem of I I've actually solved what is the maximum over all norm x L1 equals 1 of mod y transpose x so I'm given a y I'm asking what is this value and this is equal to the max magnitude entry of y which is equal to norm y infinity but by definition this is equal to the dual norm of the L1 norm so I'll write that as norm y L1 dual basically the dual norm of the L1 norm is the L infinity norm and similarly I can ask the alternative question given so in other words what I'm trying to show you here is how to find the dual norm okay given a norm we ask what is the dual norm of a given norm this is how you solve it so for given x when will equality hold and it holds when again suppose y is such that then then basically what I have to do is to choose x choose y i is equal to x i over mod x i for every i such that x i is not equal to zero and if x i equals zero it's eventually going to multiply with zero anyway but I can choose it to be zero otherwise I can choose it to be any value less than one in magnitude but because the the y infinity is bounded by one so I shouldn't choose a entry of y to be greater than one but I can choose it to be zero otherwise then this implies y infinity the dual norm which is equal to the max over all x such that x infinity equals one of mod y transpose x now if I substitute this y i into that summation above I you can see that this is equal to the mod x i will cancel with the mod x i and so you'll be left with summation of mod x i and so this is equal to norm y l1 so the dual norm of the l infinity norm is the l1 norm okay so specifically what did I do here I'll just repeat this for the sake of clarity what I did was recall that the by definition the dual norm is like this if I want to find the dual norm I want to find I want to solve this optimization problem maximize y transpose x magnitude subject to f of z equals one so I can choose this f to be some particular norm or a pre-norm and I can ask what is the dual norm in order to find this again I have to solve this optimization problem now one typical trick in solving optimization problems is to find an upper bound on the cost function and try to see if there is a z satisfying the constraint where you actually achieve this upper bound if you can find an upper bound so suppose I could find some some so if there exists some say zeta such that mod y transpose z is less than or equal to zeta for all z such that f of z equals one then if and if there exists some z prime such that mod y transpose z prime is equal to this zeta then this quantity then zeta is the dual norm so that is the that is the process I'm trying to follow here what I did first is I found an upper bound on mod y transpose x in terms of the l1 norm of x and the infinity norm of y okay and so then if I restrict the l1 norm of x to be equal to one then I know that mod of y transpose x is less than or equal to norm at the l infinity norm of y for all y then I ask can I ever at can I achieve this upper bound yes answer is yes I can achieve this upper bound of norm y infinity if I choose x such that xi equals one for the argument i which maximizes mod yk and zero otherwise and this in turn shows that the dual norm of the l1 norm is equal to the l infinity norm of y and similarly if I restrict the norm y to be y infinity equal to one then I ask whether this mod of y transpose x can ever be equal to the l1 norm of x this is the upper bound for norm y infinity equal to one and I find that the answer is yes if I just choose yi equal to xi or mod xi then if you look at what happens to y transpose x that becomes summation yi that is summation of yi xi but yi itself is xi over mod xi times xi i equal to 1 to n and this is xi times xi over mod xi is so this is equal to summation i equal to 1 to n mod xn okay so because this is xi square and I can also write that as mod xi square and that divided by xi is just mod xn and which is in turn equal to the l1 norm of x and as a consequence the dual norm of the l infinity norm is the l1 norm similarly if I take the l2 norm then I have from Cauchy Schwartz mod y transpose x is less than or equal to norm y l2 times norm x l2 and equality if and only if x and y are dependent or x is equal to some alpha times y so in particular if if y is non-zero then x choosing x to be equal to y over the l2 norm of y satisfies x l2 norm equals 1 and so so this quantity it it will and and it solves max norm x l2 equals 1 mod y transpose x so that implies the dual norm is equal to the l2 norm itself so we say that the l2 norm is its self-dual in fact it is the only norm that has this property sir in the previous one the dual norm of y infinity should be x norm of y x1 or should be y1 the previous equation dual norm of y infinity should be single one norm of x no not y no no no see when an optimization problem like this the solution to the optimization cannot contain x I've already searched over all possible x's and I'm asking what is the maximum value of this it'll only depend on y there is no meaning to writing the dual norm of y as something that depends on x axis x is like a local variable to this optimization problem it is the only norm which is self-dual another so this is also something that can be shown I won't show it here but it can be shown so if you take if you start from the property that the dual norm of a given norm is equal to the norm itself you can then derive and show that that norm must be the Euclidean norm and also another property is that the dual norm of the dual of the dual norm of a vector norm so we start with the norm we find its dual norm and then we ask what is the dual of that dual norm and you see from these two examples that if I started with the L1 norm the dual of that is the L infinity norm then if I asked what is the dual of the L infinity norm I get back the L1 norm here also in the second example anyway if I take the dual norm of the Euclidean norm I get the Euclidean norm then if I ask what is the dual of that it is again the Euclidean norm and this property is true for all norms so the dual of a of a dual norm of a vector norm is the vector norm itself you can't go on producing new norms by finding the dual of the dual of the dual and so on you can do two of them and that's it you stop there yeah what is the question yeah my question is should you take in a norm should the dual norm always exist I mean for any norm do we have a dual norm for it yes so essentially the point is you're maximizing a linear which is a convex function over a compact set so it will always have a maximizer within that compact set and so the dual norm always exists it may not always be easily computable you may need to solve an optimization problem like this in some special cases like the ones we considered it's possible to work out what is the solution to this optimization problem but that need not always be the case and so one more doubt in the LP norm that we have defined can P be any rational number or should it be only a positive integer it can be any number yeah any rational number it can even be an irrational number okay then one of the follow-up questions sir so can I so given a norm say LP norm can I say that its dual norm will be an LQ norm where P and Q satisfy the holder inequality is it necessary I think so but I need to double check that yeah so this inequality itself suggests that that this inequality is holder's inequality it's a generalization of the Cauchy-Schwarz inequality and for a given X such that XLP norm equals one if you ask what is the sorry for it excuse me for a given Y if you fix X such that XLP norm equals one and then you ask among all such vectors whose LP norm equals one when can the maximum of Y transpose X be attained and if you solve that optimization problem you will find that the maximum will be attained at an X such that the value of Y transpose X equals this upper bound which is the LQ norm of Y where Q is a number such that one over P plus one over Q equals one so those are not that easy to show algebraically but it is possible to show that and therefore the dual norm of the LP norm where P is any number between one and infinity is the LQ norm where Q is the number satisfying this inequality here one over P plus one over Q equals one. Yes okay okay that was my doubt okay thank you.